Step by step
The question
Question: find an eigenvector of A for λ = 4.
An eigenvector is a vector that A only stretches: A times v is λ times v.
The zero vector does not count.
1 Eigenvector for λ = 4
An eigenvector for this λ is a vector that A − λI sends to 0: its null space.
We take 4 off the diagonal: 3 − 4 = −1 and 0 − 4 = −4.
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − 2y = 0.
We want x alone.
To remove − 2y, we add 2y on both sides.
They cancel on the left.
So x = 2y.
We pick y = 1.
So an eigenvector for λ = 4 is (2, 1).
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
So: λ = 4: (2, 1).
Any number times them, except 0, works too.
2 Check
Let's check: A v must be λ times v.
That is 4 times (2, 1): (8, 4).
A only stretches it, by its λ.
So the answer is right!
Let's look back.
First: eigenvector for λ = 4.
Last: check.